Question

examples 1 and 2 1. triangle xyz has vertices x(0, 2), y(4, 4), and z(3, −1). graph △xyz and its image after a rotation of 180° counterclockwise about z.

Explanation:

Step1: Recall Rotation Rule

To rotate a point \((x,y)\) \(180^\circ\) counterclockwise about a center \((a,b)\), we use the rule: \((x - a, y - b)\) is rotated \(180^\circ\) to \((- (x - a), - (y - b))\), then add back \((a,b)\). So the formula for the image of \((x,y)\) after \(180^\circ\) rotation about \((a,b)\) is \((2a - x, 2b - y)\).

Step2: Find Image of X about Z

For point \(X(0,2)\) and center \(Z(3,-1)\), substitute into the formula:
\(x' = 2\times3 - 0 = 6\)
\(y' = 2\times(-1) - 2 = -2 - 2 = -4\)
So image of \(X\) is \(X'(6, -4)\).

Step3: Find Image of Y about Z

For point \(Y(4,4)\) and center \(Z(3,-1)\):
\(x' = 2\times3 - 4 = 6 - 4 = 2\)
\(y' = 2\times(-1) - 4 = -2 - 4 = -6\)
So image of \(Y\) is \(Y'(2, -6)\).

Step4: Image of Z

Since we are rotating about \(Z\), the image of \(Z\) is itself, so \(Z'(3, -1)\).

Answer:

The image of \(\triangle XYZ\) after \(180^\circ\) counterclockwise rotation about \(Z\) has vertices \(X'(6, -4)\), \(Y'(2, -6)\), and \(Z'(3, -1)\). To graph, plot the original points \(X(0,2)\), \(Y(4,4)\), \(Z(3,-1)\) and the image points \(X'(6, -4)\), \(Y'(2, -6)\), \(Z'(3, -1)\) and connect the vertices for each triangle.